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1 7 8 9 = 24: 15 solutions

Difficulty: EasyOpen in Solver

1 × 7 + 8 + 9 = 24

  1. 1 × 7 = 7
  2. 8 + 9 = 17
  3. 7 + 17 = 24

15 solutions

  • 1 × 8 + 7 + 9 = 24Steps
    1. 1 × 8 = 8
    2. 7 + 9 = 16
    3. 8 + 16 = 24
  • 1 × 9 + 7 + 8 = 24Steps
    1. 1 × 9 = 9
    2. 7 + 8 = 15
    3. 9 + 15 = 24
  • 7 ÷ 1 + 8 + 9 = 24Steps
    1. 7 ÷ 1 = 7
    2. 8 + 9 = 17
    3. 7 + 17 = 24
  • 7 + 8 + 9 ÷ 1 = 24Steps
    1. 7 + 8 = 15
    2. 9 ÷ 1 = 9
    3. 15 + 9 = 24
  • 7 + 9 + 8 ÷ 1 = 24Steps
    1. 7 + 9 = 16
    2. 8 ÷ 1 = 8
    3. 16 + 8 = 24
  • 8 × (9 + 1 − 7) = 24Steps
    1. 1 − 7 = -6
    2. 9 + -6 = 3
    3. 8 × 3 = 24
  • 9 + 1 × (7 + 8) = 24Steps
    1. 7 + 8 = 15
    2. 1 × 15 = 15
    3. 9 + 15 = 24
  • 1 × (9 + 7 + 8) = 24Steps
    1. 7 + 8 = 15
    2. 9 + 15 = 24
    3. 1 × 24 = 24
  • (9 + 7 + 8) ÷ 1 = 24Steps
    1. 7 + 8 = 15
    2. 9 + 15 = 24
    3. 24 ÷ 1 = 24
  • 9 + (7 + 8) ÷ 1 = 24Steps
    1. 7 + 8 = 15
    2. 15 ÷ 1 = 15
    3. 9 + 15 = 24
  • 8 + 1 × (7 + 9) = 24Steps
    1. 7 + 9 = 16
    2. 1 × 16 = 16
    3. 8 + 16 = 24
  • 8 + (7 + 9) ÷ 1 = 24Steps
    1. 7 + 9 = 16
    2. 16 ÷ 1 = 16
    3. 8 + 16 = 24
  • 7 + 1 × (8 + 9) = 24Steps
    1. 8 + 9 = 17
    2. 1 × 17 = 17
    3. 7 + 17 = 24
  • 7 + (8 + 9) ÷ 1 = 24Steps
    1. 8 + 9 = 17
    2. 17 ÷ 1 = 17
    3. 7 + 17 = 24

Solving hints

  • Build a factor pair: split the four numbers into two groups and try to make a factor pair of 24 — 3 × 8, 4 × 6, or 12 × 2 — then multiply the two groups.

  • No factor pair divides cleanly? Switch to sums: make two numbers add up to 24, or try shapes like (a ± b) × (c ± d) and (a + b + c) × d.

Related hands

FAQ

Can 1 7 8 9 make 24?

Yes. Every one of them is listed above with step-by-step working.

How many solutions does 1 7 8 9 have?

15 — the list above is exhaustive, so every mathematically distinct way to combine the numbers is shown.

How difficult is 1 7 8 9?

It is rated “Easy” on our four-tier scale, based on whether whole-number intermediate steps suffice, how many distinct solutions exist, and whether the hand includes a two-digit card (10–13).